Let {OA} = -4 {i} + 3 {k}, {OB} = 14 {i} + 2 {j} - 5 {k}. {OD} bisects AOB and |{OD}| = {6}, then {OD} =

Correct Answer is : ±(i‾+j‾+2k‾)± ≤ft( {i} + {j} + 2 {k} )±(i+j+2k)

± \; {1}{{2}} ≤ft( 2 {i} + {j} + {7} {k} )

Correct Answer is : ±(i‾+j‾+2k‾)± ≤ft( {i} + {j} + 2 {k} )±(i+j+2k)

Test Index

AP EAPCET 05-Jul-2022 Shift 2 Paper

Section: Mathematics

Question : 33 of 160

Marks: +1, -0

Let

\overline{OA} = -4 \overline{i} + 3 \overline{k}, \overline{OB} = 14 \overline{i} + 2 \overline{j} - 5 \overline{k}. \overline{OD}

\angle AOB

|\overline{OD}| = \sqrt{6}

\overline{OD} =

$\pm \left( \overline{i} + \overline{j} + 2 \overline{k} \right)$
$\pm \left( \overline{i} + 2 \overline{j} + \overline{k} \right)$
$\pm \left( 2 \overline{i} + \overline{j} + \overline{k} \right)$
$\pm \; \frac{1}{\sqrt{2}} \left( 2 \overline{i} + \overline{j} + \sqrt{7} \overline{k} \right)$

Solution: 👈: Video Solution

The bisector

OD

\angle AOB

\; = \lambda \left( \; \frac{OA}{|OA|} + \; \frac{OB}{|OB|} \right)

\; = \lambda \left( \; \frac{\widehat{-4 i + 3 \hat{k}}}{\sqrt{16+9}} + \; \frac{\widehat{14 i + 2 \hat{j} - 5 \hat{k}}}{\sqrt{196+4+25}} \right)

\; = \lambda \left[ \; \frac{\widehat{-4 i + 3 \hat{k}}}{5} + \; \frac{\widehat{14 i + 2 \hat{j} - 5 \hat{k}}}{15} \right]

\; = \lambda \left[ \; \frac{\widehat{-12 i + 9 \hat{k} + 14 \hat{i} + 2 \hat{j} - 5 \hat{k}}}{15} \right]

\; = \lambda \times \; \frac{2}{15} \left[ \hat{i} + \hat{j} + 2 \hat{k} \right]

\Rightarrow \; \; |OD| = \sqrt{6}

\; = \sqrt{ \left( \frac{2}{15} \lambda \right)^2 + \left( \frac{2}{15} \lambda \right)^2 + \left( \frac{4}{15} \lambda \right)^2 } = \sqrt{6}

\; = \frac{4 \lambda^2 + 4 \lambda^2 + 16 \lambda^2}{225} = 6

\; = \lambda^2 = \frac{225}{4} = \lambda = \frac{+15}{2}

\; \text{ OD } \; = \pm \frac{15}{2} \times \frac{2}{15} \left( \hat{i} + \hat{j} + 2 \hat{k} \right)

\; = \pm \left( \hat{i} + \hat{j} + 2 \hat{k} \right)

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